Question

A spring compressed by 0.10 m from its rest length launches 0.53-kg block (starting at rest) along a rough horizontal surface with the coefficient of kinetic friction 0.24. If the spring constant is 515 N/m, how far will the block slide from its initial position?

Answer 1

Answer:
`2.06\ m`

Step-by-step explanation:

Given

Spring is compressed by 0.1 m from its rest length

Mass of block is 0.53 kg

The coefficient of kinetic friction is
`\mu_k=0.24`

Spring constant
`k=515\ N/m`

Here, the spring energy is consumed against work done by friction force

Friction force acting on the mass is
`F=\mu_kmg`

Writing Elastic potential energy equals work done by friction force

`\Rightarrow (1)/(2)kx^2=\mu_kmg\cdot s\\\\\Rightarrow 0.5* 515* 0.1^2=0.24* 0.53* 9.8\cdot s\\\Rightarrow s=2.06\ m`

So, the block will slide up to a distance of
`2.06\ m`